$Ricardo$有$2020$枚硬币，其中一些是便士（$1$美分硬币），其余的是镍币（$5$美分硬币）。他有至少一枚一分便士和一枚五分镍币。$Ricardo$可能拥有的最大金额和可能拥有的最小金额相差多少美分？

A. 8062

B. 8068

C. 8072

D. 8076

E. 8082

1. 2020 AMC 8 官方解析

To obtain the greatest possible amount of money, use as many nickels as possible: $2019$ nickels and $1$ penny. This gives $2019 \cdot 5 + 1 \cdot 1 = 10,096$ cents. Similarly, to obtain the least possible amount of money, use as many pennies as possible: $1$ nickel and $2019$ pennies.
This gives $1 \cdot 5 + 2019 \cdot 1 = 2024$ cents. Therefore the difference between the greatest possible and least possible amounts of money is $10,096 - 2024 = 8072$ cents.
为了获得最大的金额，要使$5$美分硬币尽可能的多：$2019$枚$5$美分硬币和$1$枚$1$美分硬币. 金额中共为 $2019 \cdot 5 + 1 \cdot 1 = 10,096$ 美分. 同样的，要获得最小的金额，应该有尽可能多的$1$美分硬币：$1$枚$1$美元的硬币和 $2019$ 枚$5$美元硬币. 总金额为$1 \cdot 5 + 2019 \cdot 1 = 2024$美分. 因此，最大可能金额与最小可能金额之差为 $10,096 - 2024 = 8072$ 美分.
One penny and one nickel will be used in either case so they can be removed, leaving $2018$ coins. The greatest possible sum will result from having $2018$ nickels, and the least possible sum will result from having $2018$ pennies. The difference in amounts is $2018 \cdot 5 - 2018 \cdot 1 = 2018 \cdot 4 = 8072$ cents.
获得最大金额和最小金额两种情况下都会有一枚$1$美分硬币和一枚$5$美分硬币，故$2020$枚硬币中必须有一枚$1$美分硬币和一枚$5$美分硬币. 剩下 $2018$ 枚硬币. 具有 $2018$ 枚$5$美分硬币将使其获得最大的可能金额，而具有 $2018$ 枚$1$美分硬币将使其获得最小的可能金额. 故最大可能金额与最小可能金额之差为 $2018 \cdot 5 - 2018 \cdot 1 = 2018 \cdot 4 = 8072$ 美分.